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LC Circuit inductance problem

  1. Dec 13, 2009 #1
    1. The problem statement, all variables and given/known data
    A circuit consists of a capacitor and an inductor. The resistance in the circuit is small and can be neglected. Initially, at t = 0 s, the voltage across the capacitor is at its maximum of VC = 10 V, the charge stored in the capacitor is 1 mC. It is observed that the capacitor discharges to QC = 0 C after t = 0.25 ms. What is the inductance L of the inductor in the circuit?

    http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys212/oldexams/exam3/fa09/fig8.gif [Broken]

    2. Relevant equations

    Conservation of energy:

    Energy Capacitor + Energy Inductor = Qmax^2 / 2C

    3. The attempt at a solution

    Initial Energy capacitor = .5*(Qmax^2/C) = .005
    Final Energy Capacitor = 0

    Initial Energy Inductor = 0
    Final Energy Inductor = .5*L*I^2


    .005 = .5*L*I^2 where I = dQ/dt which is (1E-3)/(.25E-3) = 4

    L = 6.25E-4 H which is wrong.

    Right answer is 253.3 microH.

    What am I doing wrong?
    Thanks for any help!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 13, 2009 #2


    User Avatar

    The inductor-capacitor circuit that you describe is analogous to a simple frictionless mass-spring system. It may be described by the following differential equation:

    [tex] L \frac{d^2q}{dt^2} + \frac{1}{C}q = 0 [/tex]

    The solution is [itex] q = q_m cos(\omega t + \phi) [/itex] which gives the charge on the capacitor at any time t.

    For the initial condition you describe, the constant [itex] \phi [/itex] is zero and

    [itex] \omega = \sqrt{\frac{1}{LC}} [/itex].

    Now, the capacitor discharges in .25 msec. What is the value of [itex] \omega t [/itex] when q = 0? (Hint: it has to make the cosine = 0.) That is sufficient for you to calculate L.

    Your mistake was assuming that the current is simply q/t.
    Last edited by a moderator: May 4, 2017
  4. Dec 13, 2009 #3
    Thank you thats perfect!
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